Download Combinatorics Practice - Missouri State University

MO
MO-ARML Practice
Name
Counting Procedures
Permutations, Combinations,
Distributions and Derangements
Date
______________________________
_______________
TOPIC 1: Permutations
Example 1: The factorial 10! = 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 3 628 800.
Factorial, explicit form: In general, n! = n(n – 1)(n – 2) ... · 3 · 2 · 1,
where n is any Natural number. [ 3! is read as ‘three factorial’. ]
A permutation is any arrangement of a set of elements which is in a distinct order.
number of permutations of k members chosen from n distinct elements.
n Pk
denotes the
Example 2: In how many ways can 7 people’s names be arranged in a list of names?
Solution: 7 · 6 · 5 · 4 · 3 · 2 · 1 = 7! = 5040.
Example 3: How many permutations are there of the letters in ‘SERIES’?
Solution: There would be 6! ways, if not for the fact that the two S’s and two E’s are not
distinguishable; since it does not matter if those pairs are switched, we divide: 6! / (2! · 2!) = 180.
Example 4: In how many ways can 5-letter words be selected from the letters of ‘UNSAMPLED’?
Solution: While each letter is different, we only care about the orders formed for 5 of the 9 letters,
but not the other 4 letters; so: 9 P5 = 9! / (9 – 5)! = 9 · 8 · 7 · 6 = 3024.
Permutations, factorial form:
n
Pk =
n!
(n – k)!
, where n and k are
Natural numbers, and k ≤ n.
Example 5: If the cylinders in a six-cylinder engine are numbered from 1 to 6, then how many different
firing orders are theoretically possible?
Solution: Once a particular sequence of the 6 cylinders is selected, it will represent the same firing
order no matter which of the 6 is chosen as first, so Pc = 6 P 6 / 6 = 6! / 6 = (6 – 1)! = 120 ways.
Circular Permutations: There are
permutations of n distinct elements.
n
Pc k =
nPk
n
=
( n – 1)!
(n – k)!
circular
Problems
1) Find 9! / 6!
2) How many ways are there to list the names of five candidates for the same office on a ballot?
3) How many ways are there to scramble the letters in the word SCRAMBLE?
4) If a coach gets to choose the starting lineup positions for her intramural baseball team of four girls on
the infield and five boys in the other positions from 11 female and 8 male team members, how many
starting lineups are possible?
5) In how many ways can ten numbers be arranged around a dart board?
6) Myrtle has a small cube. Each face has been painted a different color. She wants to make the cube into
a die by putting the numbers 1 through 6 on the faces with the condition that the 1 and 6, the 2 and 5, and
the 3 and 4 must be on opposite faces. How many different ways can she mark the cube?
7) In how many ways can six differently colored beads be arranged on a bracelet?
TOPIC 2: Combinations
Example 6: In a standard 52-card deck, how many possible 5-card poker hands are there?
Solution: A representation of this might be 52 51 50 49 48, but this is not a simple pernutation.
We are interested in neither the ways the 47 cards not drawn could be arranged nor in how the five cards
are arranged in our hand. Since order does not matter at all, we divide all possible arrangements by the
arrangements of both the cards drawn and those not drawn. So, there are
52!
47!! 5!
= 2 598 960 poker hands.
All permutations which contain the same elements in a different order represent the same combination.
Combinations, factorial form: The number of combinations of n
objects taken k at a time is n C k =
Note that
nC k
n!
n!(n – k)! .
n
may also be written as C(n, k), Cn, k , C k and
( nk ) .
Pascal’s Triangle is a representation of all possible combinations:
1
0C0
1
1
1C0
1C1
1
2
1
=
2C0
2C1
2C2
1
3
3
1
3C0
3C1
3C2
3C3
1
4
6
4
1 ...
4C0
4C1
4C2
4C3
4 C 4 ...
Problems
1) How many 13-card bridge hands consist of all black cards?
2) If a coach gets to choose the starting lineup of four boys and five girls for his intramural volleyball
team from 11 male and 8 female team members, how many starting lineups are possible?.
3) How many poker hands consist of five of the same suit (called a flush)?
4) How many bridge hands have four hearts, two diamonds, and seven clubs?
5) At a pizza house, it is possible to have a pizza with any combination of the following ingredients:
pepperoni, mushrooms, peppers, sardines, sausage, anchovies, jalepeños, onions, and bacon. How many
different pizzas could one order (assume no multiples of a single ingredient)?
TOPIC 3: Distributions
Example 7: How many ordered triples (a, b, c) of non-negative integers are there so that a + b + c = 10?
Solution: We know (2, 5, 3) would be a solution, since 2 + 5 + 3 = 10. Note that this can be
represented as 11 + 11111 + 111. We see a string with 12 elements. (4, 0, 6) can be viewed as
1111 + + 111111. We need to choose the positions of two “+” signs from 12 positions. Hence, we need
!
$
12!
12 '11
12
to compute 12 C 2 . #" 2 &% = 10! 2! = ! 2
= 66.
In classical language, we would state:
The number of ways b indistinguishable balls can be distributed into u distinguishable urns is
& b + (u ' 1) #
$$
!!. .
% (u ' 1) "
Problems
1) A bakery offers four kinds of cookies – sugar, chocolate chip, vanilla, and oatmeal. A young lady
plans to buy just four cookies. How many combinations are there? Think of a string with separators.
2) Five identical balls are distributed into three distinct jars marked A, B, and C. Find the number of
distributions.
3) Find the number of ways to place five red balls and five black balls into three jars marked A, B, and C.
4) Find the number of ways to distribute five balls into three distinct boxes if each box contains at least
one ball.
5) Find the number of distributions of 16 balls into jars A, B, and C if A has at least 3, B at least 4, and C
has exactly 2.
TOPIC 4: Derangements
A derangement of an ordered sequence, a1, a2, a3, … , an, leaves no element in its original position.
Dn denotes the number of derangements of n distinct elements in a sequence.
Example 8: 35412 is a derangement of 12345. (The sequence is viewed as a string of digits.)
Example 9: How many derangements are there of a sequence of three elements?
Solution: There are two derangements of 1, 2, 3. They are 3, 1, 2 and 2, 3, 1. We write D3 = 2.
Problems
1) D1 =
2) D2 =
3) List the derangements of a, b, c, d. What is the value of D4?
Upon examining D5 = 44 and D6 = 265, a recursive pattern emerges. D6 = 6·D5 + (-1)6 =
6(44) + 1 = 264 + 1 = 265.
4) D4 = 4D3 + (-1)4 =
5) D7 =
Derangements, Recursive Form:
Dn = nDn-1 + (-1)n
6) Calculate 7!( 1 – 1/1! + 1/2! – 1/3! + 1/4! – 1/5! + 1/6! – 1/7!)
Derangements, Factorial Form:
Dn = n!(1 –
1
1!
!
1
2!
+
1
3!
!
1
+ ...
4!
+ (-1)n "
1
n! ).
7) Find the number of rearrangements of 1234 such that 1 is not in position 3, 2 is not in position 1, 3 is
not in position 2, and 4 is not in position 4.
8) Find the number of derangements of 12345 such that the first element is 2.